Question

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find $\frac{dy}{dx}$ for $y = \frac{\tan x}{sin^{2}x}$.
$\frac{dy}{dx}=square$

Explanation:

Step1: Simplify the function

First, rewrite $\tan x=\frac{\sin x}{\cos x}$, so $y = \frac{\tan x}{\sin^{2}x}=\frac{\frac{\sin x}{\cos x}}{\sin^{2}x}=\frac{1}{\sin x\cos x}=\csc x\sec x$.

Step2: Use the product - rule

The product - rule states that if $y = u\cdot v$, where $u=\csc x$ and $v = \sec x$, then $\frac{dy}{dx}=u'v + uv'$.
We know that $u'=-\csc x\cot x$ and $v'=\sec x\tan x$.

Step3: Calculate the derivative

$\frac{dy}{dx}=-\csc x\cot x\sec x+\csc x\sec x\tan x$.
Since $\cot x=\frac{\cos x}{\sin x}$ and $\tan x=\frac{\sin x}{\cos x}$, we have:
\[

$$\begin{align*} \frac{dy}{dx}&=-\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\cdot\frac{1}{\cos x}+\frac{1}{\sin x}\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\\ &=-\frac{1}{\sin^{2}x}+\frac{1}{\cos^{2}x}\\ &=\sec^{2}x-\csc^{2}x \end{align*}$$

\]

Answer:

$\sec^{2}x - \csc^{2}x$